Metamath Proof Explorer


Theorem ackval42

Description: The Ackermann function at (4,2). (Contributed by AV, 9-May-2024)

Ref Expression
Assertion ackval42 ⊢ Ack ⁡ 4 ⁡ 2 = 2 65536 − 3

Proof

Step Hyp Ref Expression
1 df-4 ⊢ 4 = 3 + 1
2 1 fveq2i ⊢ Ack ⁡ 4 = Ack ⁡ 3 + 1
3 df-2 ⊢ 2 = 1 + 1
4 2 3 fveq12i ⊢ Ack ⁡ 4 ⁡ 2 = Ack ⁡ 3 + 1 ⁡ 1 + 1
5 3nn0 ⊢ 3 ∈ ℕ 0
6 1nn0 ⊢ 1 ∈ ℕ 0
7 ackvalsucsucval ⊢ 3 ∈ ℕ 0 ∧ 1 ∈ ℕ 0 → Ack ⁡ 3 + 1 ⁡ 1 + 1 = Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 1
8 5 6 7 mp2an ⊢ Ack ⁡ 3 + 1 ⁡ 1 + 1 = Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 1
9 3p1e4 ⊢ 3 + 1 = 4
10 9 fveq2i ⊢ Ack ⁡ 3 + 1 = Ack ⁡ 4
11 10 fveq1i ⊢ Ack ⁡ 3 + 1 ⁡ 1 = Ack ⁡ 4 ⁡ 1
12 ackval41a ⊢ Ack ⁡ 4 ⁡ 1 = 2 16 − 3
13 11 12 eqtri ⊢ Ack ⁡ 3 + 1 ⁡ 1 = 2 16 − 3
14 13 fveq2i ⊢ Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 1 = Ack ⁡ 3 ⁡ 2 16 − 3
15 2cn ⊢ 2 ∈ ℂ
16 6nn0 ⊢ 6 ∈ ℕ 0
17 6 16 deccl ⊢ 16 ∈ ℕ 0
18 expcl ⊢ 2 ∈ ℂ ∧ 16 ∈ ℕ 0 → 2 16 ∈ ℂ
19 15 17 18 mp2an ⊢ 2 16 ∈ ℂ
20 3cn ⊢ 3 ∈ ℂ
21 ackval3 ⊢ Ack ⁡ 3 = n ∈ ℕ 0 ⟼ 2 n + 3 − 3
22 oveq1 ⊢ n = 2 16 − 3 → n + 3 = 2 16 - 3 + 3
23 npcan ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ → 2 16 - 3 + 3 = 2 16
24 22 23 sylan9eqr ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ ∧ n = 2 16 − 3 → n + 3 = 2 16
25 24 oveq2d ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ ∧ n = 2 16 − 3 → 2 n + 3 = 2 2 16
26 25 oveq1d ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ ∧ n = 2 16 − 3 → 2 n + 3 − 3 = 2 2 16 − 3
27 3re ⊢ 3 ∈ ℝ
28 4re ⊢ 4 ∈ ℝ
29 3lt4 ⊢ 3 < 4
30 27 28 29 ltleii ⊢ 3 ≤ 4
31 sq2 ⊢ 2 2 = 4
32 30 31 breqtrri ⊢ 3 ≤ 2 2
33 2re ⊢ 2 ∈ ℝ
34 1le2 ⊢ 1 ≤ 2
35 17 nn0zi ⊢ 16 ∈ ℤ
36 1nn ⊢ 1 ∈ ℕ
37 2nn0 ⊢ 2 ∈ ℕ 0
38 9re ⊢ 9 ∈ ℝ
39 2lt9 ⊢ 2 < 9
40 33 38 39 ltleii ⊢ 2 ≤ 9
41 36 16 37 40 declei ⊢ 2 ≤ 16
42 2z ⊢ 2 ∈ ℤ
43 42 eluz1i ⊢ 16 ∈ ℤ ≥ 2 ↔ 16 ∈ ℤ ∧ 2 ≤ 16
44 35 41 43 mpbir2an ⊢ 16 ∈ ℤ ≥ 2
45 leexp2a ⊢ 2 ∈ ℝ ∧ 1 ≤ 2 ∧ 16 ∈ ℤ ≥ 2 → 2 2 ≤ 2 16
46 33 34 44 45 mp3an ⊢ 2 2 ≤ 2 16
47 4nn0 ⊢ 4 ∈ ℕ 0
48 31 47 eqeltri ⊢ 2 2 ∈ ℕ 0
49 48 nn0rei ⊢ 2 2 ∈ ℝ
50 37 17 nn0expcli ⊢ 2 16 ∈ ℕ 0
51 50 nn0rei ⊢ 2 16 ∈ ℝ
52 27 49 51 letri ⊢ 3 ≤ 2 2 ∧ 2 2 ≤ 2 16 → 3 ≤ 2 16
53 32 46 52 mp2an ⊢ 3 ≤ 2 16
54 nn0sub ⊢ 3 ∈ ℕ 0 ∧ 2 16 ∈ ℕ 0 → 3 ≤ 2 16 ↔ 2 16 − 3 ∈ ℕ 0
55 5 50 54 mp2an ⊢ 3 ≤ 2 16 ↔ 2 16 − 3 ∈ ℕ 0
56 53 55 mpbi ⊢ 2 16 − 3 ∈ ℕ 0
57 56 a1i ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ → 2 16 − 3 ∈ ℕ 0
58 ovexd ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ → 2 2 16 − 3 ∈ V
59 21 26 57 58 fvmptd2 ⊢ 2 16 ∈ ℂ ∧ 3 ∈ ℂ → Ack ⁡ 3 ⁡ 2 16 − 3 = 2 2 16 − 3
60 19 20 59 mp2an ⊢ Ack ⁡ 3 ⁡ 2 16 − 3 = 2 2 16 − 3
61 2exp16 ⊢ 2 16 = 65536
62 61 oveq2i ⊢ 2 2 16 = 2 65536
63 62 oveq1i ⊢ 2 2 16 − 3 = 2 65536 − 3
64 14 60 63 3eqtri ⊢ Ack ⁡ 3 ⁡ Ack ⁡ 3 + 1 ⁡ 1 = 2 65536 − 3
65 4 8 64 3eqtri ⊢ Ack ⁡ 4 ⁡ 2 = 2 65536 − 3